php
  • mysql
  • database
  • fatal-error
  • url-parsing
  • 2017-07-01 34 views 1 likes 
    1

    試圖從PHP試圖從PHP

    $conn = mysqli_connect($servername, $username, $password, $dbname); 
    
    // Find status for row with Username of the Url?username= 
    $sql = "SELECT Status FROM Users WHERE Username = '".$_GET["username"]."'"; 
    $result = mysql_query($sql) or die('Error connecting to database'); 
    $username = mysql_result($result, 0, "Status"); 
    
    echo 'Username Status is ' . $Status; 
    mysqli_close($conn); 
    
    
    ?> 
    

    在結果我得到讀取我的MySQL數據庫中的用戶的狀態讀取我的MySQL DATABSE用戶的狀態是這樣的:

    Fatal error: Uncaught Error: Call to undefined function mysql_query() in /storage/ssd4/269/2113269/public_html/teststat.php:11 Stack trace: #0 {main} thrown in /storage/ssd4/269/2113269/public_html/teststat.php on line 11

    回答

    0

    你的語法mysqli是不正確的,因爲mysqli不同於mysql語法。

    MySQLi代表MySQL改進。它是一個面向對象的接口,可以讓MySQL的綁定更容易使用。

    我建議你使用的mysqli版本,並沒有查詢股價的MySQL語法與mysqli的語法。

    $servername = "localhost"; 
    $username = "username"; 
    $password = "password"; 
    $dbname = "myDB"; 
    
    // Create connection 
    $conn = new mysqli($servername, $username, $password, $dbname); 
    // Check connection 
    if ($conn->connect_error) { 
        die("Connection failed: " . $conn->connect_error); 
    } 
    
    $sql = "SELECT Status FROM Users WHERE Username = '".$_GET["username"]."'"; 
    $result = $conn->query($sql); 
    
    if ($result->num_rows > 0) { 
        // output data of each row or remove while loop if you wish 
        while($row = $result->fetch_assoc()) { 
         echo 'Username Status is ' . $row['Status'] ; 
        } 
    } else { 
        echo "0 results"; 
    } 
    $conn->close(); 
    
    +0

    完美謝謝!這是因爲在試圖使這個腳本更好地工作時,我從一箇舊網站(它使用mysql)複製。 –

    +0

    @Luke Deven,很高興它適合你。 :) –

    0

    你混了mysql和mysqli的

    $result = mysql_query($sql) or die('Error connecting to database'); 
    $username = mysql_result($result, 0, "Status"); 
    
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